std::remquo, std::remquof, std::remquol
| Defined in header <cmath>
|
||
| (1) | ||
float remquo ( float x, float y, int* quo ); double remquo ( double x, double y, int* quo ); |
(since C++11) (until C++23) |
|
| constexpr /* floating-point-type */ remquo ( /* floating-point-type */ x, |
(since C++23) | |
| float remquof( float x, float y, int* quo ); |
(2) | (since C++11) (constexpr since C++23) |
| long double remquol( long double x, long double y, int* quo ); |
(3) | (since C++11) (constexpr since C++23) |
| Defined in header <cmath>
|
||
| template< class Arithmetic1, class Arithmetic2 > /* common-floating-point-type */ |
(A) | (since C++11) (constexpr since C++23) |
std::remquo for all cv-unqualified floating-point types as the type of the parameters x and y.(since C++23)Contents |
[edit] Parameters
| x, y | - | floating-point or integer values |
| quo | - | pointer to int to store the sign and some bits of x / y |
[edit] Return value
If successful, returns the floating-point remainder of the division x / y as defined in std::remainder, and stores, in *quo, the sign and at least three of the least significant bits of x / y (formally, stores a value whose sign is the sign of x / y and whose magnitude is congruent modulo 2n to the magnitude of the integral quotient of x / y, where n is an implementation-defined integer greater than or equal to 3).
If y is zero, the value stored in *quo is unspecified.
If a domain error occurs, an implementation-defined value is returned (NaN where supported).
If a range error occurs due to underflow, the correct result is returned if subnormals are supported.
If y is zero, but the domain error does not occur, zero is returned.
[edit] Error handling
Errors are reported as specified in math_errhandling.
Domain error may occur if y is zero.
If the implementation supports IEEE floating-point arithmetic (IEC 60559),
- The current rounding mode has no effect.
- FE_INEXACT is never raised.
- If x is ±∞ and y is not NaN, NaN is returned and FE_INVALID is raised.
- If y is ±0 and x is not NaN, NaN is returned and FE_INVALID is raised.
- If either x or y is NaN, NaN is returned.
[edit] Notes
POSIX requires that a domain error occurs if x is infinite or y is zero.
This function is useful when implementing periodic functions with the period exactly representable as a floating-point value: when calculating sin(πx) for a very large x, calling std::sin directly may result in a large error, but if the function argument is first reduced with std::remquo, the low-order bits of the quotient may be used to determine the sign and the octant of the result within the period, while the remainder may be used to calculate the value with high precision.
On some platforms this operation is supported by hardware (and, for example, on Intel CPUs, FPREM1 leaves exactly 3 bits of precision in the quotient when complete).
The additional overloads are not required to be provided exactly as (A). They only need to be sufficient to ensure that for their first argument num1 and second argument num2:
|
(until C++23) |
|
If num1 and num2 have arithmetic types, then std::remquo(num1, num2, quo) has the same effect as std::remquo(static_cast</* common-floating-point-type */>(num1), If no such floating-point type with the greatest rank and subrank exists, then overload resolution does not result in a usable candidate from the overloads provided. |
(since C++23) |
[edit] Example
#include <cfenv> #include <cmath> #include <iostream> #ifndef __GNUC__ #pragma STDC FENV_ACCESS ON #endif const double pi = std::acos(-1); // or std::numbers::pi since C++20 double cos_pi_x_naive(double x) { return std::cos(pi * x); } // the period is 2, values are (0;0.5) positive, (0.5;1.5) negative, (1.5,2) positive double cos_pi_x_smart(double x) { int quadrant; double rem = std::remquo(x, 1, &quadrant); quadrant = static_cast<unsigned>(quadrant) % 2; // The period is 2. return quadrant == 0 ? std::cos(pi * rem) : -std::cos(pi * rem); } int main() { std::cout << std::showpos << "naive:\n" << " cos(pi * 0.25) = " << cos_pi_x_naive(0.25) << '\n' << " cos(pi * 1.25) = " << cos_pi_x_naive(1.25) << '\n' << " cos(pi * 2.25) = " << cos_pi_x_naive(2.25) << '\n' << "smart:\n" << " cos(pi * 0.25) = " << cos_pi_x_smart(0.25) << '\n' << " cos(pi * 1.25) = " << cos_pi_x_smart(1.25) << '\n' << " cos(pi * 2.25) = " << cos_pi_x_smart(2.25) << '\n' << "naive:\n" << " cos(pi * 1000000000000.25) = " << cos_pi_x_naive(1000000000000.25) << '\n' << " cos(pi * 1000000000001.25) = " << cos_pi_x_naive(1000000000001.25) << '\n' << "smart:\n" << " cos(pi * 1000000000000.25) = " << cos_pi_x_smart(1000000000000.25) << '\n' << " cos(pi * 1000000000001.25) = " << cos_pi_x_smart(1000000000001.25) << '\n'; // error handling std::feclearexcept(FE_ALL_EXCEPT); int quo; std::cout << "remquo(+Inf, 1) = " << std::remquo(INFINITY, 1, &quo) << '\n'; if (fetestexcept(FE_INVALID)) std::cout << " FE_INVALID raised\n"; }
Possible output:
naive: cos(pi * 0.25) = +0.707107 cos(pi * 1.25) = -0.707107 cos(pi * 2.25) = +0.707107 smart: cos(pi * 0.25) = +0.707107 cos(pi * 1.25) = -0.707107 cos(pi * 2.25) = +0.707107 naive: cos(pi * 1000000000000.25) = +0.707123 cos(pi * 1000000000001.25) = -0.707117 smart: cos(pi * 1000000000000.25) = +0.707107 cos(pi * 1000000000001.25) = -0.707107 remquo(+Inf, 1) = -nan FE_INVALID raised
[edit] See also
| (C++11) |
computes quotient and remainder of integer division (function) |
| (C++11)(C++11) |
remainder of the floating point division operation (function) |
| (C++11)(C++11)(C++11) |
signed remainder of the division operation (function) |
| C documentation for remquo
| |
